Question: A curve in the plane is defined parametrically by the equations $x=3-4t^2$ and $y=4e^{6t}$. Find the value of $\dfrac{dy}{dx}$ at $t=1$. Choose 1 answer: Choose 1 answer: (Choice A) A $-\dfrac{e^6}{2}$ (Choice B) B $-2e^6$ (Choice C) C $-3e^6$ (Choice D) D $4e^6$
Explanation: In general, to find the derivative (i.e. the expression for $\dfrac{dy}{dx}$ ) of a function defined parametrically by the equations $x=u(t)$ and $y=v(t)$ (where $u$ and $v$ are any functions of $t$ ), we use the following rule: $\dfrac{dy}{dx}=\dfrac{\left(\dfrac{dy}{dt}\right)}{\left(\dfrac{dx}{dt}\right)}=\dfrac{v'(t)}{u'(t)}$ We are given that $x=3-4t^2$ and $y=4e^{6t}$ : $\begin{aligned} \dfrac{dy}{dx}&=\dfrac{\dfrac{d}{dt}\left(4e^{6t}\right)}{\dfrac{d}{dt}(3-4t^2)} \\\\ &=\dfrac{24e^{6t}}{-8t} \\\\ &=\dfrac{-3e^{6t}}{t} \gray{\text{Simplify}} \end{aligned}$ Now let's evaluate $\dfrac{dy}{dx}$ at $t= 1$ : $\begin{aligned} &\phantom{=}\dfrac{-3e^{6( 1)}}{ 1} \\\\ &=-3e^6 \end{aligned}$ In conclusion, the value of $\dfrac{dy}{dx}$ at $t=1$ is $-3e^6$.